This paper describes exactly what you need.
Read this, and then this.
Then if you read a little deeper, you'll find this:
Here is some EulerAngles Code, which you can have a look at...
this google search also gives lots of other solutions to your problem.
You can actually use these methods to recover the regular rotation angles too, as constructing the rotation matrix is trivial, and you only need to reverse the order of the euler rotations to get the the regular rotation angles.
(where by regular and euler, i just mean intrinsic and extrinsic)
EDIT: just realised i answered my own question by accident...
The classic consecutive matrix product $R_2 R_1$ applies to rotations about original axes (extrinsic).
Normally, it is more easy to visualize , e.g., a rotation around $z$, followed by a rotation around the new axis $y'$ (intrinsic).
But $y'=R_z \, y$, and to convert the rotation $R_{y'}$ around $y'$ into a rotation around $y$, you shall first revert $R_z$, apply the rotation around $y$, bring again $y$ to $y'$, so we can write
$$R_{y'}R_z = (R_z R_y {R_z}^{-1}) R_z = R_z R_y$$
And that is the same for additional rotations.
That corresponds to how a linear map transforms under a change of the reference system: what $R_{y'}$ produces in the new system is what $(R_z R_y {R_z}^{-1})$ does in the original one.
In fact, if we have a rotation (or any linear transform) that applied to a generic vector $\bf v$ provides a vector $\bf w$
$$
{\bf w} = {\bf R}\,{\bf v}
$$
if we transform the two vectors through a linear invertible matrix $\bf T$ (so including rotations, but not only), i.e.
$$
{\bf v'} = {\bf T}\,{\bf v}\quad {\bf w'} = {\bf T}\,{\bf w}
$$
then we have
$$
{\bf w'} = {\bf T}\,{\bf w} = {\bf T}\,{\bf R}\,{\bf v} = {\bf T}\,{\bf R}\,{\bf T}^{\, - \,{\bf 1}} \;{\bf Tv} = \left( {{\bf T}\,{\bf R}\,{\bf T}^{\, - \,{\bf 1}} } \right){\bf v'}
$$
The matrix $\bf R$ and ${{\bf T}\,{\bf R}\,{\bf T}^{\, - \,{\bf 1}} }$ are similar matrices
Best Answer
No, Euler angles are a poor choice for representing the orientation. Apart from that not all orientations can be represented it complicates the calculations.
What you instead could do is to use quaternion representation of orientation or use a rotation matrix. That would make the maths more easy, the relation between angular velocity and orientation is $\omega q = {1\over 2}{d\over dt}\operatorname{vec}(q)$, where $\operatorname{vec}(q)$ is the vector part of the quaternion.